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$$\lim_{y\to 0} \frac{y}{\cos\left(\frac{\pi}{2}(1+y)\right)}$$ Can anybody help me? I can use basic properties of limits, and some of those basic known limits. I know it would be easier with derivatives, but I was just wondering if it's possible without L-Hospital's rule, derivatives, Taylor series.

Thank you in advance!

My ideas for now: changing cosine into sine. Maybe that. I have no other clue.

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2 Answers 2

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HINT:

$$\cos\left(\dfrac\pi2+A\right)=-\sin A$$

and $$\lim_{h\to0}\dfrac{\sin h}h=\text{ ?}$$

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Note that $$\lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}(1+y)\right)} = \lim_{y\to 0}\frac{y}{\cos\left(\frac{\pi}{2}+\frac{\pi y}{2}\right)} = \lim_{y\to 0}\frac{y}{-\sin\left(\frac{\pi y}{2}\right)} \\ = -\frac{2}{\pi}\lim_{y\to 0}\frac{\left(\frac{\pi y}{2}\right)}{\sin\left(\frac{\pi y}{2}\right)} = -\frac{2}{\pi}(1) = -\frac{2}{\pi}$$

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  • 1
    $\begingroup$ Thank you for your help :) $\endgroup$
    – MathIsTheWayOfLife
    Commented Jul 31, 2015 at 12:47
  • $\begingroup$ How come we are assuming 0 / 0 is 1, or even that the limit approaches 1. . ? $\endgroup$
    – Andrew
    Commented Jul 31, 2015 at 23:58
  • $\begingroup$ @Lemony: We aren't: we're using one of the "basic known limits" involving $\sin$. $\endgroup$
    – user14972
    Commented Aug 1, 2015 at 1:27
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    $\begingroup$ @Hukyl: absolutely right. $\endgroup$
    – Harish Chandra Rajpoot
    Commented Aug 1, 2015 at 1:30
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    $\begingroup$ @user246608: $\lim\limits_{x\to0}\frac{\sin(x)}x=1$ is derived geometrically in this answer. $\endgroup$
    – robjohn ♦
    Commented Aug 1, 2015 at 4:34

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